Question: Let $h(x)=x^{^{\scriptsize\dfrac{1}{4}}}$. $h'(16)=$
Let's first find the expression for $h'(x)$ and then evaluate it at $x=16$. The derivative of $h$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{4}}}\right) \\\\ &=\dfrac{1}{4}x^{^{\frac{1}{4}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac{1}{4}x^{^{-\frac{3}{4}}} \end{aligned}$ So we found that $h'(x)=\dfrac{1}{4}x^{^{-\frac{3}{4}}}$, which can also be written as $\dfrac{1}{4(\sqrt[4]{x})^3}$ Now let's plug ${x=16}$ : $\begin{aligned} \dfrac{1}{4(\sqrt[4]{{16}})^3}&=\dfrac{1}{4(2)^3} \\\\ &=\dfrac{1}{4\cdot 8} \\\\ &=\dfrac{1}{32} \end{aligned}$ In conclusion, $h'(16)=\dfrac{1}{32}$.